f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
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f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
-1<=sin(2x+2A+π/6)<=1
所以-1<=2sin(2x+2A+π/6)+1<=3
f(x)=-a+1
所以有解的-1<=-a+1<=3
-3<=a-1<=1
-2<=a<=2
f(x)-m=2sin(2x-π/3) 1-m=2 2sin(2x-π/3)=m 2-1=m 1 x∈【π/4,π/2】 2x-π/3∈【π/6,π/2】 1≤2sin(2x-π/3)≤2
f(x)+a-1=0
即2sin(2x+2A+π/6)+1+a-1=2sin(2x+2A+π/6)+a=0 有解
即a=-2sin(2x+2A+π/6) 所以a∈【-2,2】
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